Question

sinθ.sin(60 - θ).sin(60 + θ) = 1/4 sin 3θ .

Answer 1

L. H. S
=> Sinθ.sin(60 - θ).sin(60 + θ)
=>sinθ[sin60cosθ-sinθcos60][sin60cosθ+sinθcos60}
=>sinθ[(sin60cosθ)²-(sinθcos60)²]

=>sinθ[sin²60cos²θ- sin²θcos²60]
=>sinθ[3/4cos²θ-sin²θ*1/4]
=>sinθ*1/4[3cos²θ-sin²θ]




=>sinθ/4[3(1-sin²θ)-sin²θ]
=>sinθ/4[3-3sin²θ-sin²θ]
=>1/4*sinθ[3-4sin²θ]
=>1/4[3sinθ-4sin³θ]
=>1/4sin 3θ